Product of Factorials modulo Prime

Relevant wiki: Wilson's Theorem

Note first that $n \equiv -(101 - n) \pmod{101}$ for $1 \le n \le 51$ implies that

$51! \equiv (-50)(-51)(-52)...(-99)(-100) \pmod{101} \equiv (-1)^{51}\dfrac{100!}{49!} \pmod{101}$ .

Thus $49!51! \equiv (-1)^{51}100! \pmod{101} \equiv (-1)(-1) \pmod{101} \equiv \boxed{1} \pmod{101}$ ,

since by Wilson's Theorem $(p - 1)! \equiv -1 \pmod{p}$ for any prime $p$ .